cos(a-b)=-12/13,cos(a+b)=12/13,a-b(π/2,π),a+b(3π/2,2π)cos2a,cos2b,b=?

问题描述:

cos(a-b)=-12/13,cos(a+b)=12/13,a-b(π/2,π),a+b(3π/2,2π)cos2a,cos2b,b=?

a-b(π/2,π),a+b(3π/2,2π) cos(a-b)=-12/13,cos(a+b)=12/13
sin(a-b)=5/13 sin(a+b)=-5/13 2b(π/2,3π/2)
cos=cos(a-b+a+b)= cos(a-b)cos(a+b)-sin(a-b) sin(a+b)=119/169
cos2b=cos(-2b)=c0s(a-b-(a+b))=1
b=π/2